Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;

public class Main {

    Scanner sc = new Scanner(System.in);

    public static void main(String[] args) {
        new Main().run();
    }

    void run() {

        int n = sc.nextInt();
        double[] p = new double[n];
        for (int i=0; i<n; i++) p[i] = sc.nextDouble();

        double[][] dp = new double[n+1][n+1];

        dp[0][0] = 1;

        for (int i=1; i<=n; i++) {
            for (int j=0; j<=n; j++) {
                if (j == 0) {
                    dp[i][j] = dp[i-1][j] * (1 - p[i-1]);
                    continue;
                }
                dp[i][j] = dp[i-1][j] * (1 - p[i-1]) + dp[i-1][j-1] * p[i-1];
            }
        }

        double res = 0;

        for (int i=n/2+1; i<=n; i++) {
            res += dp[n][i];
        }

//        System.out.println(dp[1][1]);
//        System.out.println(dp[1][0]);

        System.out.println(res);

    }

}